Let [tex]a[/tex] be the acceleration of the masses. By Newton's second law, we have
• for the masses on the left,
[tex]1.3mg - T = 1.3ma[/tex]
where [tex]T[/tex] is the magnitude of tension in the pulley cord, and
• for the mass on the right,
[tex]T - mg = ma[/tex]
Eliminate [tex]T[/tex] to get
[tex](1.3mg - T) + (T - mg) = 1.3ma + ma[/tex]
[tex]0.3mg = 2.3ma[/tex]
[tex]\implies a = \dfrac{0.3}{2.3}g \approx 0.13g \approx 1.3 \dfrac{\rm m}{\mathrm s^2}[/tex]
Starting from rest and accelerating uniformly, the right-hand mass moves up 75 cm = 0.75 m and attains an upward velocity [tex]v[/tex] such that
[tex]v^2 = 2a(0.75\,\mathrm m) \\\\ \implies v \approx \sqrt{2\left(1.3\frac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx 1.4\dfrac{\rm m}{\rm s}[/tex]
When the 0.5m mass is released, the new net force equations change to
• for the mass on the right,
[tex]mg - T' = ma'[/tex]
where [tex]T'[/tex] and [tex]a'[/tex] are still tension and acceleration, but not having the same magnitude as before the mass was removed; and
• for the mass on the left,
[tex]T' - 0.8mg = 0.8ma'[/tex]
Eliminate [tex]T'[/tex].
[tex](mg - T') + (T' - 0.8mg) = ma' + 0.8ma'[/tex]
[tex]0.2mg = 1.8 ma'[/tex]
[tex]\implies a' = \dfrac{0.2}{1.8}g = \dfrac19 g \approx 1.1\dfrac{\rm m}{\mathrm s^2}[/tex]
Now, the right-hand mass has an initial upward velocity of [tex]v[/tex], but we're now treating down as the positive direction. As it returns to its starting position, its speed [tex]v'[/tex] at that point is such that
[tex]{v'}^2 - v^2 = 2a'(0.75\,\mathrm m) \\\\ \implies v' \approx \sqrt{\left(1.4\dfrac{\rm m}{\rm s}\right)^2 + 2\left(1.1\dfrac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx \boxed{1.9\dfrac{\rm m}{\rm s}}[/tex]
